# Wonders of math

1. Dec 19, 2004

### mahesh_2961

hi ,
i recently found while playing around with my new calculator that any number repeated six times is divisible by 7 ....
for eg, 4 repeated 6 times gives 444444/7 =63492
an say 761 repeated 6 times 761761761761761761/ 7 =108823108823108823
the last one was done using a computer ...
can anyone plz help me out why this happens???

regrds
Mahesh

2. Dec 19, 2004

### mahesh_2961

one correction to my query ... not any number but any integer
sorry for that

Mahesh

3. Dec 19, 2004

### Zurtex

227153715052227153715052227153715052227153715052227153715052227153715052 is 227153715052 repeated 6 times and is not a multiple of 7 as its prime factors are:

2 ^ 2 x 3 x 11 x 19 x 37 x 67 x 73 x 137 x 3169 x 52579 x 98641 x
333667 x 2082527 x 99990001 x 999999000001 x 3199044596370769

Edit: Sorry got a little carried away on my computer

Last edited: Dec 19, 2004
4. Dec 19, 2004

### mahesh_2961

hi ,
repeating an integer, x (say), six times is not the same as multiplying it six times ....
i m sorry but i think u havent got my question correctly...
it is like this
suppose x is an n digit number
repeating it 6 times gives a number
y= 1*x + 10^(n)*x + 10^(2n)*x + 10^(3n)*x+ ....+10^(6n)*x
which is not equal to x^6
and i think i have got the solution for this problem....
if u want me to post it plz tell me , or if u want to find it urself then its good for u

regards
Mahesh

5. Dec 19, 2004

### gonzo

well, it's easy to prove for any one digit number n:

10^5*n+10^4*n+10^3*n+10^2*n+10*n+n

That is a single digit n repeated 6 times, divide by 7 and you get:

14285 5/7 * n +
1428 4/7 * n +
142 6/7 * n +
14 2/7 * n +
1 3/7 * n +
1/7 * n =

15870*n + (21/7)*n = 15873 * n

So, a single digit repeated 6 times will always divide by 7 to an integer. In other words, if you multiply 15873 * 7 * n , where n is a single digit, then you will get a number that is that digit repeated 6 times.

If you feel like it, you can attempt similar proofs for more than one digit to see if it holds for more than one digit repeated, and if not than maybe you can find certain conditions for which it would hold with bigger numbers.

Or maybe someone has a better way to prove this than me ...

6. Dec 19, 2004

### Muzza

It does not appear to be true for 123423123423123423123423123423123423 either (that is 123423 repeated 6 times).

Last edited: Dec 19, 2004
7. Dec 19, 2004

### matt grime

Repeating a string, s, of r digits six times is the sam as the sum of a geometric progression. Let k be 10^r, it is

s +ks +k^2s+..+k^5s

The sum of this is (s/9)*(10^6-1)

since 9=2 mod 7, and thus invertible, and 10 is coprime with 7, this reduces to zero mod 7 quite trivially by fermat's little theorem.

8. Dec 19, 2004

### gonzo

You do realize this post doesn't make much sense, do you? First, the person you responded to didn't multiple a number 6 times, he/she did what you did, and just repeated one.

And your formula is wrong. It should only go up to 10^(5n).

9. Dec 19, 2004

### gonzo

Matt, could you explain this snippet a bit more for me? I've done some modular arithmatic to determine factorability, but this loses me. Thanks.

10. Dec 19, 2004

### Zurtex

227153715052227153715052227153715052227153715052227153715052227153715052 is 227153715052 repeated 6 times and not (227153715052)6 and it is not a multiple of 7.

11. Dec 19, 2004

### matt grime

Either you or I have made a mistake, then, Zurtex. Now, if there truly were counter examples, then there would be a smaller on than the one you've just written. Incidentally, what is the remainder of that number mod 7?

As for dividing by 9.

suppose x/9 = y in the integers, then x=9y, and x=9y mod any integer, n say. If 9 and n are relativeyl prime then there is an integer k such that 9k=1 mod n thus xk=y mod n. 9 is a unit modulo 9, 9 is invertible. If 9 and 7 weren't coprime then I couldn't do this.

12. Dec 19, 2004

### gonzo

okay, I followed everything you said individually, but I don't understand "invertible".

I'm missing some piece of the puzzle here. Sorry for being dense. Maybe you could write it out in few steps starting from

s/9 * (10^6 -1)

I would start like this myself (which seems to be more complicated and possibly wrong):

10^6=10^3*10^3 and 10^3 mod 7 = -1

so the whole thing mod 7 = s/9(-1*-1 -1) = 0

13. Dec 19, 2004

### shmoe

Your geometric series should be s/(k-1)*(k^6-1)

k-1 won't be invertible mod 7 if r=0 mod 6 (Fermat's little theorem). Note Zurtex example has r=12 and Muzza has r=6. r=6 is the least number of digits you'll find for a counterexample.

Last edited: Dec 19, 2004
14. Dec 19, 2004

### matt grime

ah, there we go, for some reason i'd got 9 rather than 10^r - 1.

15. Dec 19, 2004

### Ethereal

I think mahesh really meant the following as he himself corrected:

But in any case, what range of number string lengths can be repeated six times and be divisible by 7?

16. Dec 19, 2004

### dextercioby

I believe Shmoe's post was pretty clear.Basically any natural number except natural multiples of 6.

Daniel.

17. Dec 19, 2004

### gonzo

Is this because 10^6 mod 7 = 1 so that k-1=0?

Can someone give me a really simple definition of "invertible" here? Thanks.

18. Dec 19, 2004

### Zurtex

I've not been able to keep up the thread but I think you're in the wrong matt:

227153715052227153715052227153715052227153715052227153715052227153715052 mod 7 = 2

There will be smaller counter examples but that's the 1st one I found (mind you I only searched for about 10 secs.

19. Dec 19, 2004

### mahesh_2961

hai ,
u ppl were right.... when i found one of the solutions for the problem ... i thought it was quite an easy one ... and u rightly pointed out the mistake i made in the formula ...
i apologize for anything from my side...

This cannot be called a proof....
i noted that (1/7) = 0.142857142857142857..... with '142857' reccuring so
y= 1*x + 10^(n)*x + 10^(2n)*x + 10^(3n)*x+ 10^(4n)*x +10^(5n)*x
=x(1+10^n+10^2n+10^3n+10^4n+10^5n)
Now i multiplied both sides with 0.142857142857.....
and found that RHS is always an integer irrespective of the values of x and hence n.
thank u guys for helping me

cheers
Mahesh

20. Dec 19, 2004

### matt grime

yep. i am, and shmoe has pointed out where my error is.

21. Dec 19, 2004

### Galileo

Muzza found 123423 repeated six times.
It factors into $(3)^2(19)(101)(999999000001)(333667)(52579)(41141)(9901)$

Repeating 123422 gives a smaller one

22. Dec 19, 2004

### Zurtex

Good good but I was the 1st reply (other than the original poster) with this counter example, there was no need for all of this nonsense

23. Dec 19, 2004

### Muzza

And 100000 repeating smaller yet... ;)

24. Dec 19, 2004

### matt grime

at least we have established a sufficient condition for it to be true though (however not necessary).

25. Dec 19, 2004

### shmoe

That's correct.

we say s is invertible mod n if there exists a t with st=1 mod n. You can show that s is invertible mod n if and only if s and n have no common factors except 1 (they're relatively prime). In this case you can find the inverse by using the Euclidean algorithm to write 1 as a linear combination of s and n. Thus for 9 mod 7:

9=7*1+2
7=2*3+1

Turning this around we get:

1=7-2*3=7*4+(-3)*9

And we see (-3)*9=4*9=1 mod 7

so 9 is invertible mod 7 and it's inverse is 4. Of course you could have found this inverse by inspection easily enough, but the euclidean algortihm will work for the general case.